复数数列 $\{w_n\}$ 的定义如下： $w_n = \dfrac{[1+(n-1)i][1+(n+1)i]}{(1+ni)^2}\text{，其中 }n\in\mathbb{Z}^+.$

1. Apply the properties of arguments to split the expression.

$$\begin{align*} \arg(w_n) &= \arg\left( \dfrac{[1+(n-1)i][1+(n+1)i]}{(1+ni)^2} \right) \\ &= \arg([1+(n-1)i][1+(n+1)i]) - \arg((1+ni)^2) \\ &= \arg(1+(n-1)i) + \arg(1+(n+1)i) - 2\arg(1+ni) \end{align*}$$

By comparing this to $\arg(w_n) = p[\arg(1+(n-1)i)] + q[\arg(1+ni)] + r[\arg(1+(n+1)i)]$, we have $p=1$, $q=-2$ and $r=1$.

1. Express $\arg(z_n)$ as a sum of $\arg(w_k)$ and substitute the result from part (i).

From part (i), using the property $\arg\left(\dfrac{ab}{c^2}\right) = \arg(a) + \arg(b) - 2\arg(c)$, we have:
$\arg(w_k) = \arg(1+(k-1)i) - 2\arg(1+ki) + \arg(1+(k+1)i)$

Since $z_n = w_1 w_2 \dots w_n$, by the properties of arguments of products:
$\arg(z_n) = \sum_{k=1}^n \arg(w_k)$

2. List the first few and last few terms of the series to show the pattern of differences.

We write out the terms of the sum to observe the cancellation pattern:

$$\begin{array}{r l} k=1: & \arg(1) - 2\arg(1+i) + \arg(1+2i) \\ k=2: & \arg(1+i) - 2\arg(1+2i) + \arg(1+3i) \\ k=3: & \arg(1+2i) - 2\arg(1+3i) + \arg(1+4i) \\ \vdots & \\ k=n-1: & \arg(1+(n-2)i) - 2\arg(1+(n-1)i) + \arg(1+ni) \\ k=n: & \arg(1+(n-1)i) - 2\arg(1+ni) + \arg(1+(n+1)i) \end{array}$$

3. Perform the cancellation to obtain the simplified expression for the sum.

Summing the columns vertically, most terms cancel out.
For $2 \le m \le n-1$, the term $\arg(1+mi)$ appears with coefficients $+1, -2,$ and $+1$, summing to $0$.

The remaining uncancelled terms are:
$\arg(z_n) = \arg(1) - \arg(1+i) - \arg(1+ni) + \arg(1+(n+1)i)$

4. Evaluate the constant arguments to complete the proof.

Evaluate the constant arguments:
$\arg(1) = 0$
$\arg(1+i) = \tan^{-1}\left(\frac{1}{1}\right) = \frac{1}{4}\pi$

Substituting these values into the sum gives:
$\arg(z_n) = 0 - \frac{1}{4}\pi - \arg(1+ni) + \arg(1+(n+1)i)$
$\arg(z_n) = -\frac{1}{4}\pi - \arg(1+ni) + \arg[1+(n+1)i]$ (shown)

1. Evaluate the limit of $\arg(z_n)$ as $n \to \infty$ by converting the arguments to inverse tangents.

From part (ii), we have $\arg(z_n) = -\dfrac{\pi}{4} - \arg(1+ni) + \arg(1+(n+1)i)$.

Writing the arguments in terms of inverse tangent:
$\arg(1+ni) = \tan^{-1}(n)$
$\arg(1+(n+1)i) = \tan^{-1}(n+1)$

As $n \to \infty$, both $\tan^{-1}(n)$ and $\tan^{-1}(n+1)$ approach $\dfrac{\pi}{2}$.

$\begin{aligned} \lim_{n \to \infty} \arg(z_n) &= -\frac{\pi}{4} - \lim_{n \to \infty} \tan^{-1}(n) + \lim_{n \to \infty} \tan^{-1}(n+1) \\ &= -\frac{\pi}{4} - \frac{\pi}{2} + \frac{\pi}{2} \\ &= -\frac{\pi}{4} \end{aligned}$

2. Determine the linear relationship between the real and imaginary parts using the limiting argument.

The argument of a complex number $z = x + iy$ determines the ratio of its imaginary part to its real part via $\tan(\theta) = \dfrac{y}{x}$.

As $n \to \infty$, $\arg(z_n) \to -\dfrac{\pi}{4}$. Therefore:
$\frac{\text{Im}(z_n)}{\text{Re}(z_n)} \to \tan\left(-\frac{\pi}{4}\right) = -1$

Thus, the linear relationship as $n \to \infty$ is:
$\text{Im}(z_n) = -\text{Re}(z_n)$